Pipework Networks (Hardy cross method)

[GT90]

Hi everyone, I have a 2nd year fluid mechanics question!

Homework Statement

I'm having trouble gripping what this question is looking for:

Explain and Discuss the rationale implications of assuming that the head loss in each section of the pipe network is proportional to the square of the flow rate?

Homework Equations

head loss equation, H(loss) = R * m^2
where R=resistance in the pipe and m=mass flow rateIf you know fluid mechanics any help would be greatly appreciated!
Thanks

 

[Valkarie]

!The head loss in any given section of a pipe is often assumed to be proportional to the square of the flow rate. This assumption implies that the resistance in the pipe is also proportional to the square of the flow rate, where R is the resistance in the pipe and m is the mass flow rate. This is based on the fact that the pressure drop (or head loss) across the pipe is proportional to the square of the velocity of the fluid, according to the Darcy-Weisbach equation.The rationale behind this assumption is that it helps simplify the analysis of a pipe network by allowing us to estimate the head losses in each section of the pipe without having to directly measure the actual flow velocity. This makes it easier to calculate the total head loss in the system, which can then be used to estimate the pressure drop across the entire network. Moreover, this assumption also allows us to predict the pressure drop in each section of the pipe network, depending on the flow rate of the fluid. In short, assuming that the head loss in each section of the pipe network is proportional to the square of the flow rate simplifies the analysis of a pipe network, and provides an easy way to calculate the total head loss and pressure drop across the entire system.